Aptitude Discussion

Q. |
Sachin can cover a distance in 1hr 24 min by covering 2/3 of the distance at 4 kmph and the rest at 5 kmph. The total distance is? |

✖ A. |
5 km |

✔ B. |
6 km |

✖ C. |
7 km |

✖ D. |
8 km |

**Solution:**

Option(**B**) is correct

Let total distance $= D$

Distance travelled at 4 kmph speed $= \left(\dfrac{2}{3}\right)D$

Distance travelled at 5 kmph speed $= \left(1-\dfrac{2}{3}\right)D = \left(\dfrac{1}{3}\right)D$

Total time =1 hr 24 min = (60+24) min $=\dfrac{84}{60}$ hr $= \dfrac{21}{15}$ hr

We know,

\(\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}\)

Total time

\(\dfrac{21}{15}=\dfrac{2/3}{4}D+\dfrac{1/3}{5}D\)

\(\dfrac{21}{15}=\dfrac{2D}{12}+\dfrac{D}{15}\)

\(84=14D\)

$D= \textbf{6 km}$

**Nilay**

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**Parth Singh**

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Thanks to all for clearing my doubts...

**Pankaj Sharma**

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time for total distance cover=7/5 hr

if total distance 3 then it cover 2 km by the speed 4 kmph then he will take 1/2 hr

rest mean 1 km distance cover by 5 kmph speed then he will take 1/5 hr

now total time = 1/2 +1/5 = 7/10---*2----> 7/5 hr

so total distance will be 3-----*2------> 6 km

**Anonymous**

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I didn't understand how the 21/15 come.Please help me

Total time given in the question is 1 hr 24 min. When it is converted into hours, it becomes $\dfrac{21}{15}$.

This is how it is done:

$1 \text{ hr } 24 \text{ min }= 60+24 \text{ min}$

$=84 \text{ min} = \dfrac{84}{60}\text{ hr}$

$=\dfrac{21}{15} \text{ hr}$

**Madhu**

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Thanks to Rohan and Deepti. Both of you clear my doubt.

**Madhu**

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Thank you for your reply.After seeing next question to this I understand how 1/3 comes but don't understand why they subtract 2/3 from 1. From where the 1 comes?Dothey assume the distance as 1?

No Madhu,

Total distance here is not 1. Question mention that journey is devided into 2 parts, first part being $\dfrac{2}{4}$. To find out the remaining part we subtract this distance from the total distance (complete journey is 1 PART).

Basically instead of dealing with absolute distances we are dealing here with the ratios.

One example clears everything. Let me give you an example.

This is all about ratios, Total distance may be anything.

Let us assume the total distance to be 15 km, thus 1st part becomes $=\dfrac{2}{3} \times 15= 10$ km.

Similarly remaining distance would be:

$=\left(1-\dfrac{2}{3}\right)\times 15$

$=\left(\dfrac{1}{3}\right)\times 15=5$ km.

**Madhu**

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I didn't understand how the 1/3 come.Please help me.

Question says that Sachin can cover a distance in 1hr 24 min by covering 2/3 of the distance at 4 kmph and the rest at 5 kmph. Here REST of the distance is $\left(1-\dfrac{2}{3}\right)=\dfrac{1}{3}$.

This is how the factor of $\dfrac{1}{3}$ comes into picture.

T1+T2=(7/5)

=> (D1/S1)+(D2/S2)=(7/5)

=> ((2x/3)/4)+((1x/3)/5)=(7/5)

=> ((14x/3)/20)=(7/5)

=> x=6 kms